How many different signals can be made with 5 different flags placed side by side if two particular flags must always be next to each other?

Last updated at Dec. 24, 2019 by Teachoo

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Ex7.1, 6 Given 5 flags of different colours, how many different signals can be generated if each signal requires the use of 2 flags, one below the other? A signal can have only 2 flags The required number of signals = 5 × 4 = 20

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How many signals can be made with the aid of $5$ flags of different colors all arranged in a row?

blancoangelica227 blancoangelica227

Answer:

20 signals

Step-by-step explanation:

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lumikidmaryrose lumikidmaryrose

Answer:

20 signals

Step-by-step explanation:

sana makatulong

Answer

Verified

Hint: We are given with the fact that the total number of flags are 5. Then, the signals can be generated 1,2,3,4 and 5 flags from the 5 flags. Thus then we use permutation and find the number of ways to arrange the flags in 5 cases and we add them to get the total number of different signals.

Complete step-by-step answer:

We have been told that there are 5 different flags of different colors. Thus, the total number of flags=5.Now we have been told to find the number of different signals that can be generated by arranging any number of flags. Also, we have been told that "any number of flags" means that we can take 1,2,3,4 or5 flags to generate the required signal and we need to find all the cases.Then, by alteration in the arrangement of flags, the signal will change. So, we have to find the permutation which is the arrangement of the flags and we are at liberty to use any number of signals.Thus, we get the following cases as:Case 1: Using 1 flag out of 5 flags to generate the signal. Hence according to permutation, the number of permutation of 5 flags taken 1 at a time is given as, where n = 5 and r=1 \[{}^{n}{{P}_{r}}\,\,=\,\,\,{}^{5}{{P}_{1}}\]Now, by using the formula for permutation as\[{}^{n}{{P}_{r}}\,=\,\dfrac{n!}{\left( n-r \right)!}\], we get:\[\begin{align} & {}^{5}{{P}_{1}}\,\,=\,\dfrac{5!}{\left( 5-1 \right)!}\, \\ & \Rightarrow {}^{5}{{P}_{1}}=\,\dfrac{5!}{4!}\, \\ & \Rightarrow {}^{5}{{P}_{1}}\,=\,\,\dfrac{5!\,x\,4!}{4!}\,\, \\ & \Rightarrow {}^{5}{{P}_{1}}=\,\,5 \\ \end{align}\]So, we get number of different signals by using 1 flag=5…….(1)Case 2: Using 2 flags out of 5 flags to generate the signal$={}^{5}{{P}_{2}}$\[\begin{align} & {}^{5}{{\text{P}}_{2}}\text{ = }\dfrac{5!}{(5-2)!} \\ & \Rightarrow {}^{5}{{\text{P}}_{2}}\text{= 5}\times \text{4 } \\ & \Rightarrow {}^{5}{{\text{P}}_{2}}\text{= 20 } \\ \end{align}\]So, we get number of different signals by using 2 flags=20……..(2)Case3: Using 3 flags out of 5 flags to generate the signal$={}^{5}{{P}_{3}}$$\begin{align} & {}^{5}{{P}_{3}}\text{=}\dfrac{5!}{(5-3)!} \\ & \Rightarrow {}^{5}{{P}_{3}}\text{=5}\times \text{4}\times \text{3} \\ & \Rightarrow {}^{5}{{P}_{3}}\text{=60} \\ \end{align}$So, number of different signals generated by using 3 flags=60…….(3)Case 4: Using 4 flags out of 5 flags to generate the signal$={}^{5}{{P}_{4}}$$\begin{align} & {}^{5}{{P}_{4}}\text{=}\dfrac{5!}{(5-4)!} \\ & \Rightarrow {}^{5}{{P}_{4}}\text{= 5}\times \text{4}\times \text{3}\times \text{2} \\ & \Rightarrow {}^{5}{{P}_{4}}\text{=120} \\ \end{align}$So, we get number of different signal generated by using 4 flags = 120……(4)Case5: Using all 5 flags to generate the signal$={}^{5}{{P}_{5}}$.$\begin{align} & {}^{5}{{P}_{5}}\text{=}\dfrac{5!}{(5-5)!} \\ & \Rightarrow {}^{5}{{P}_{5}}\text{= 5}\times \text{4}\times \text{3}\times \text{2}\times \text{1} \\ & \Rightarrow {}^{5}{{P}_{5}}\text{=120} \\ \end{align}$So, we get number of different signal generated by using all 5 flags=120…..(5) Hence, the required number of signals is equal to sum of (1), (2), (3), (4) & (5)So, Required no. of signals as:\[\text{5}+\text{2}0+\text{6}0+\text{12}0+\text{12}0=325\]Hence, the number of different signals generated are 325 signals.

Note: We can solve this without using permutation, without repetition number of signals generated using 1 flag out of 5 flags, 5 ways number of signals generated using 2 flags out of 5 flags, 5x4= 20 ways

3 flags out of 5 flags \[=\text{5 x 4 x 3 }=\text{6}0\]ways4 flags out of 5 flags \[=\text{5 x 4 x 3 x 2}=\text{12}0\]waysAll of 5 flags \[=\text{5 x 4 x 3 x 2 x 1 }=\text{12}0\]waysSo, Required number of signals \[=\text{5}+\text{2}0+\text{6}0\text{ }+\text{12}0+\text{12}0=\text{325}\]